(N/A) Given,
Radius of the balloon,$r = 10 \, m$
$\therefore$ Volume of the balloon $= \frac{4}{3} \pi r^{3} = \frac{4}{3} \times 3.14159 \times 10^{3} = 4188.79 \, m^{3}$.
Thus,the volume of the displaced air is $4188.79 \, m^{3}$.
Given,Density of air $= 1.2 \, kg \, m^{-3}$.
Mass of displaced air $= 4188.79 \times 1.2 = 5026.55 \, kg$.
Now,mass of helium $(m)$ inside the balloon is given by $m = \frac{M p V}{R T}$.
Here,$M = 4 \times 10^{-3} \, kg \, mol^{-1}$,$p = 1.66 \, bar$,$V = 4188.79 \, m^{3} = 4188.79 \times 10^{3} \, dm^{3}$,$R = 0.083 \, bar \, dm^{3} \, K^{-1} \, mol^{-1}$,$T = 300 \, K$.
$m = \frac{4 \times 10^{-3} \times 1.66 \times 4188.79 \times 10^{3}}{0.083 \times 300} = 1117.01 \, kg$.
Total mass of the balloon filled with helium $= (100 + 1117.01) \, kg = 1217.01 \, kg$.
Pay load $= (5026.55 - 1217.01) \, kg = 3809.54 \, kg$.